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NEET Sample Paper NEET Sample Test Paper-74

  • question_answer
    Two identical rods MN and OP, each of length L, cross-sectional area A and thermal conductivity K are connected as shown in figure. Ends M, O and P are maintained at temperatures\[{{T}_{1}} =\,\,20{}^\circ C\], \[{{\operatorname{T}}_{2}}=30{}^\circ C\,\,and\,\,{{T}_{3}}\,\,=\,\,40{}^\circ C\] respectively. The temperature at N is

    A)  \[32{}^\circ \text{ }C\]                        

    B)  \[33{}^\circ \text{ }C\]

    C)  \[34{}^\circ \text{ }C\]                        

    D)  \[35{}^\circ \text{ }C\]

    Correct Answer: A

    Solution :

    Let \[{{T}_{N}}\] be the temperature at N. The rate of flow of heat from O towards N is \[\frac{{{Q}_{1}}}{t}=\frac{KA({{T}_{2}}-{{T}_{N}})}{L/2}\] The rate of flow of heat from P towards N \[\frac{{{Q}_{2}}}{t}=\frac{KA({{T}_{3}}-{{T}_{N}})}{L/2}\] The rate of flow of heat from N towards M is \[\frac{{{Q}_{3}}}{t}=\frac{KA({{T}_{N}}-{{T}_{1}})}{L}\] In the steady state, (the rate at which heat enter at \[\operatorname{N}=rate at which heat leavesN)\] i.e.,        \[\frac{{{Q}_{1}}}{t}+\frac{{{Q}_{2}}}{t}=\frac{{{Q}_{3}}}{t}\] \[\frac{2KA({{T}_{2}}-{{T}_{N}})}{L}+\frac{2KA({{T}_{3}}-{{T}_{N}})}{L}=\frac{KA({{T}_{N}}-{{T}_{1}})}{L}\] \[or\,\,\,2({{T}_{2}}-{{T}_{N}})\,\,+\,\,2({{T}_{3}}-{{T}_{N}})=({{T}_{N}}-{{T}_{1}})\] Which gives \[{{T}_{N}}=\frac{{{T}_{1}}+2{{T}_{2}}+2{{T}_{3}}}{5}\] \[=\,\,\,\frac{20+2\times 30+2\times 40}{5}=32{}^\circ \,C\]


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